Math

Quote:

What are the solutions of the equation x^2-3x-4=0

this one can be factored pretty easily by inspection

x ²-3x-4 = (x-4)(x+1)

so your solutions are x = 4 and x = -1

if you're no good with factoring by eye you can always use the trusty quadratic equation which says if

ax ² + bx + c = 0

then the solutions are given by

x = [size=200]([/size]-b +/- sqrt[size=150]([/size]b ² - 4ac[size=150])[/size][size=200])[/size][size=200]/[/size][size=150]([/size]2a[size=150])[/size]

in this case we've got a=1, b=(-3), c=(-4)

plugging and chugging we get

x = (3 +/- sqrt(9+16))/2 = (3 +/- 5)/2 = 8/2 and -2/2 = 4 and -1 which agree with what we've got above

kaylinfrost:

What are the solutions of the equation x^2-3x-4=0

This is how you factorise (s if you are Auatralian, z if you are from the US)

You have to find 2 numbers that multiply to -4 and add to -3
Since they mult to a negative number, one is neg and one is pos
since the add to a neg the bigger one is the neg
-4,1 They mult ot -4 and add to -3, that's what you want

(x+1)(x-4) = 0

now, if you multiply 2 things together and get 0 then one of them must have be zero in the first place.
That is
x+1=0 OR x-4=0
x=-1 OR x=4